Question

The molar specific heat of an ideal gas at constant pressure and constant volume is '\(C_P\)' and '\(C_V\)' respectively. If '\(R\)' is a universal gas constant and the ratio of '\(C_P\)' to '\(C_V\)' is \(\gamma\), then '\(C_P\)' is equal to

• A. \((\frac{\gamma-1}{\gamma+1})R\)
• B. \(\frac{(\gamma-1)R}{\gamma}\)
• C. \(\frac{R\gamma}{(\gamma-1)}\)
• D. \(\frac{R\gamma}{(\gamma+1)}\)

Hint:

Use two key relations: \(\gamma = C_P/C_V\) and \(C_P - C_V = R\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For an ideal gas, the difference between molar specific heats is equal to the gas constant \(R\) (Mayer's relation).
The ratio of specific heats (\(\gamma\)) is a characteristic property of the gas related to its atomicity.
Step 2: Key Formula or Approach:
Mayer's Relation: \(C_P - C_V = R\)
Given ratio: \(\frac{C_P}{C_V} = \gamma \implies C_V = \frac{C_P}{\gamma}\)
Step 3: Detailed Explanation:
Substitute the expression for \(C_V\) into Mayer's relation: \[ C_P - \frac{C_P}{\gamma} = R \] Factor out \(C_P\): \[ C_P \left( 1 - \frac{1}{\gamma} \right) = R \] Simplify the term inside the parenthesis: \[ C_P \left( \frac{\gamma - 1}{\gamma} \right) = R \] Solve for \(C_P\): \[ C_P = \frac{R\gamma}{\gamma - 1} \] Similarly, we can find \(C_V = \frac{R}{\gamma - 1}\).
Step 4: Final Answer:
The molar specific heat at constant pressure is \(C_P = \frac{R\gamma}{\gamma - 1}\).